Class B power amplifier efficiency

A class B power amplifier biases each transistor at cut-off, so each conducts for 180° (half of the cycle). Two complementary transistors work in push-pull: the NPN amplifies the positive half-cycle and the PNP the negative half.

Circuit diagram (complementary-symmetry class B)

              +VCC
               |
            C /
   Vi o--+--B  Q1 (NPN)
         |    E \
         |       +-----||----+
         |       |           |
         |    E /          [RL]
         +--B  Q2 (PNP)      |
            C \             GND
               |
              -VCC

Derivation of efficiency

Let the load current be sinusoidal with peak Im (and peak output voltage Vm = Im RL).

Output (ac) power:

Po = Vm Im/2 = Im² RL/2 = Vm²/(2RL)

Input (dc) power: Each transistor draws a half-sine current pulse from its supply. The average of a half-sine of peak Im is Im/π. Both supplies together:

Idc = 2Im/π, Pi = VCC · Idc = 2 VCC Im/π

Efficiency:

η = Po/Pi = (Im² RL/2) / (2 VCC Im/π) = π Im RL/(4 VCC) = (π/4)(Vm/VCC)

So η = 78.5 % × (Vm/VCC) (general expression η = (π/4)(Vm/VCC)).

Maximum efficiency

The largest undistorted peak output is Vm = VCC (ignoring VCE(sat)):

ηmax = π/4 = 0.785 → 78.5 %

Remarks

• Power dissipated in the two transistors: P2Q = Pi − Po. It is maximum at Vm = 2VCC/π, not at full swing, where P2Q(max) = 2VCC²/(π² RL).
• Class B is much more efficient than class A (25 % or 50 %), but has crossover distortion.